Inverting Consequences of Uniform Convergence

[r.a.c.]

Hi. Now you probably know that if a function f_k(x) converges uniformly to f(x) then we are allowed to certain actions such as

lim_{n-> \inf} \int f (of k) dx = \int f dx

In other words we are allowed to exchange limit and integral. Now say we have any sequnce valued function f_k(x) . And we also have

lim_{n-> \inf} \int f (of k) dx = 0

Does that imply that f_k(x) converges uniformly to 0?

 

[Landau]

Please learn Latex properly.

Indeed, if (f_k)_{k\in\mathbb{N}} converges uniformly to f on some interval [a,b]\subset\mathbb{R}, then \lim_{k\to\infty}\int_a^b f_k(x)dx=\int_a^b \lim_{k\to\infty}f_k(x)dx=\int_a^b f(x)dx.

The converse of this is:

if \lim_{k\to\infty}\int_a^b f_k(x)dx=\int_a^b f(x)dx, does it follow that (f_k)_{k\in\mathbb{N}} converges uniformly to f?

This is not what you asked, after all \int_a^b f(x)dx=0 does not imply f=0 (i.e. f(x)=0 for all x).

The converse is therefore obviously not true. Take for example [a,b]=[0,1], and define f_k and f on [0,1] by f_k(x)=1/k and f(x)=-1 if x<1/2, f(x)=+1 if x>1/2. Then

\lim_{k\to\infty}\int_0^1 f_k(x)dx=\lim_{k\to\infty}\frac{1}{k}=0=-\frac{1}{2}+\frac{1}{2}=\int_0^1 f(x)dx

but (f_k)_k does not even converge to f pointwise.

For an counter-example to your claim, just take (f_k)_k to be a constant sequence: f_k=f as defined above, for all k. Then the integral of f_k is equal to zero (and hence the limit of the integral too), but f_k does not converge (even pointwise) to the zero function.

 

[r.a.c.]

Thanks a lot. And will do.

 

[Landau]

You're welcome. You can just click on any equation to see its Latex code.